METRIC AND TOPOLOGICAL SPACES
Part IB of the Mathematical Tripos of Cambridge
This course, consisting of 12 hours of lectures, was given by Prof. Pelham Wilson in Easter
Term 2012. This LATEX version of the notes was prepared by Henry Mak, last revised in June
2012, and is available online at http://people.pwf.cam.ac.uk/hwhm3/. Comments and corrections
to hwhm3@cam.ac.uk. No part of this document may be used for profit.
Course schedules
Metrics: Definition and examples. Limits and continuity. Open sets and neighbourhoods.
Characterizing limits and continuity using neighbourhoods and open sets.
[3]
Topology: Definition of a topology. Metric topologies. Further examples. Neighbourhoods,
closed sets, convergence and continuity. Hausdorff spaces. Homeomorphisms. Topological and
non-topological properties. Completeness. Subspace, quotient and product topologies.
[3]
Connectedness: Definition using open sets and integer-valued functions. Examples, including intervals. Components. The continuous image of a connected space is connected.
Path-connectedness. Path-connected spaces are connected but not conversely. Connected open
sets in Euclidean space are path-connected.
[3]
Compactness: Definition using open covers. Examples: finite sets and Œ0; 1. Closed subsets
of compact spaces are compact. Compact subsets of a Hausdorff space must be closed. The
compact subsets of the real line. Continuous images of compact sets are compact. Quotient spaces.
Continuous real-valued functions on a compact space are bounded and attain their bounds. The
product of two compact spaces is compact. The compact subsets of Euclidean space. Sequential
compactness.
[3]
1
M E T R I C S PA C E S
Contents
2
1
1.1
1.2
1.3
1.4
5
7
9
9
2
14
3
25
§1
Introduction
Interiors and closures
Base of open subsets for a topology
Subspaces, quotients and products
Connectedness
3.1 Various results
3.2 Path-connectedness
3.3 Products of connected spaces
19
20
21
Introduction
Open balls and open sets
Limits and continuity
Completeness
Topological spaces
2.1
2.2
2.3
2.4
13
16
Metric spaces
4
Compactness
4.1 Various results
4.2 Sequential compactness
M ETRIC SPACES
1.1 Introduction
Consider the Euclidean space Rn equipped with the standard Euclidean
inner product: Given
P
x; y 2 Rn with coordinates xi ; yi respectively, we define .x; y/ ´ niD1 xi yi , sometimes denoted
by the dot product x y.
From this we have the Euclidean norm on Rn , kxk ´ .x; x/ =2 , representing the length of
1=2
P
the vector x. We have a distance function d2 .x; y/ ´ kx yk D
yi /2 , very often
i .xi
written as d simply. This is an example of a metric.
1
Definition 1.1. A metric space .X; d / consists of a set X and a function,
called the metric, d W X X ! R such that, for all P; Q; R 2 X :
(i) d.P; Q/ > 0, with equality iff P D Q;
(ii) d.P; Q/ D d.Q; P /;
(iii) d.P; Q/ C d.Q; R/ > d.P; R/.
Condition (iii) is called the triangle inequality.
With the Euclidean metric, for any (possibly degenerate) triangle with vertices P; Q; R, the
sum of the lengths of two sides of the triangle is at least the length of the third side.
2
1.1
INTRODUCTION
Proposition 1.2. The Euclidean distance function d2 on Rn is a metric in the
sense of Definition 1.1.
Proof. (i) and (ii) are immediate. For (iii), use Cauchy-Schwarz inequality which says
n
X
!2
xi yi
6
i D1
n
X
n
X
!
xi2
i D1
!
yi2 ;
i D1
or in inner product notation, .x; y/2 6 kxk2 kyk2 for x; y 2 Rn . (See Lemma 1.3)
R
y
xCy
Q
x
P
We take P to be the origin, Q with position vector x with respect to P , and R with position
vector y with respect to Q. Then R has position vector x C y with respect to P . Now
kx C yk2 D kxk2 C 2.x; y/ C kyk2
6 kxk2 C 2kxk kyk C kyk2 D .kxk C kyk/2 :
This implies d.P; R/ D kx C yk 6 kxk C kyk D d.P; Q/ C d.Q; R/.
Lemma 1.3. (Cauchy-Schwarz). .x; y/2 6 kxk2 kyk2 for any x; y 2 Rn .
Proof. For x ¤ 0, the quadratic polynomial in the real variable kx C yk2 D 2 kxk2 C 2.x; y/ C kyk2
is non-negative for all . Considering the discriminant, we have 4.x; y/2 6 4kxk2 kyk2 .
Remarks.
1.
2.
In the Euclidean case, equality in the triangle inequality , Q lies on the straight
line segment PR. (See Example Sheet 1, Question 2)
The argument for Cauchy-Schwarz above generalises to integrals. For example,
if f; g are continuous functions on Œ0; 1, consider
Z
1
Z
2
.f C g/ > 0 )
0
fg
0
3
2
1
Z
1
f
6
0
2
Z
0
1
g2:
1.1
INTRODUCTION
More examples of metric spaces
(i)
Let X ´ Rn , and d1 .x; y/ ´
are both metrics.
Pn
i D1
jxi
yi j or d1 .x; y/ ´ maxi jxi
yi j. These
(ii) Let X be any set, and for x; y 2 X , define the discrete metric to be
(
1 if x ¤ y;
ddisc .x; y/ ´
0 if x D y:
(iii) Let X ´ C Œ0; 1 ´ ff W Œ0; 1 ! R where f is continuousg. We can define
metrics d1 , d2 , d1 on X by
Z 1
d1 .f; g/ ´
jf gj;
0
Z
d2 .f; g/ ´
1
.f
g/2
21
;
0
d1 .f; g/ ´ sup jf .x/
g.x/j:
x2Œ0;1
For d2 , the triangle inequality follows from Cauchy-Schwarz for integrals, i.e.
2
R 2 R 2
fg 6
f
g . See Remark 2 after Lemma 1.3, and use the same
argument as in Proposition 1.2.
R
(iv) British rail metric. Consider Rn with the Euclidean metric d , and let O denote the
origin. Define a new metric on Rn by
(
d.P; O/ C d.O; Q/ if P ¤ Q;
.P; Q/ ´
0
if P D Q;
i.e. all journeys from P to Q ¤ P go via O. (All rail journeys go via London.)
Some metrics in fact satisfy a stronger triangle inequality. A metric space .X; d / is called
ultra-metric if d satisfies condition (iii)0 :
d.P; R/ 6 max fd.P; Q/; d.Q; R/g
for all P; Q; R 2 X.
Example. Let X ´ Z and p be a prime. The p-adic metric is defined by
(
0
if m D n;
dp .m; n/ ´
r
1=p if m ¤ n where r D max fs 2 N W p s j .m
n/g:
We claim that d is an ultra metric. Indeed, suppose that dp .m; n/ D 1=p r1 and
dp .n; q/ D 1=p r2 for distinct m; n; q 2 Z. Then p r1 j .m n/ and p r2 j .n q/ together
imply p min fr1 ;r2 g j .m q/. So, for some r > min fr1 ; r2 g,
dp .m; q/ D 1=p r 6 1=p min fr1 ;r2 g
D max f1=p r1 ; 1=p r2 g
D max fdp .m; n/; dp .n; q/g:
4
1.2
OPEN BALLS AND OPEN SETS
This extends to a p-adic metric on Q: For any rational x ¤ y, we can write x y D
p m=n, where r 2 Z, m; n are coprime to p, and define dp .x; y/ D 1=p r similarly. Then
we also have .Q; dp / as an ultra-metric space.
r
Example. The sequence an ´ 1 C p C C p n 1 , where p is prime, is convergent in
.Q; dp / with limit a ´ .1 p/ 1 . This is because dp .an ; a/ D 1=p n for all n, and so
dp .an ; a/ ! 0 as n ! 1.
Lipschitz equivalence
Definition 1.4. Two metrics 1 ; 2 on a set X are Lipschitz equivalent if
90 < 1 6 2 2 R such that 1 1 6 2 6 2 1 .
p
Remark. For metrics d1 ; d2 ; d1 on Rn , one can show that d1 > d2 > d1 > d2 = n >
d1 =n, so they are all Lipschitz equivalent. (See Example Sheet 1, Question 9)
Proposition 1.5. d1 ; d1 on C Œ0; 1 are not Lipschitz equivalent.
Proof. For n > 1, let fn 2 C Œ0; 1 be as follows:
p
n
fn
0
1
n
2
n
1
p
p
d1 .fn ; 0/ D Area of the triangle D 1= n ! 0 as n ! 1, while d1 .fn ; 0/ D n ! 1 as
n ! 1.
Exercise. Show that d2 .fn ; 0/ D
either d1 or d1 on C Œ0; 1.
q
2
3
for all n, and so d2 is not Lipschitz equivalent to
1.2 Open balls and open sets
Let .X; d / be a metric space, and let P 2 X , ı > 0. We define the open ball by Bd .P; ı/ ´
fQ 2 X W d.P; Q/ < ıg. This is often written as B.P; ı/ or Bı .P / simply.
Examples.
1.
2.
3.
4.
In .R; d1 /, we get open intervals of the form .P ı; P C ı/.
In .R2 ; d2 /, we get open discs of radius ı; With .R2 ; d1 / we get squares; With
.R2 ; d1 /, we get tilted squares. (See Figures 1 (i), (ii) and (iii) below.)
In .C Œ0; 1; d1 /, see the example in Figure 1 (iv) below.
In .X; ddisc / where X is any set, B P; 21 D fP g for all P 2 X.
5
1.2
OPEN BALLS AND OPEN SETS
P
P
ı
P
ı
(i)
ı
(ii)
(iii)
ı
f
g
0
1
(iv)
Figure 1: B.P; ı/ in (i) .R2 ; d2 /, (ii) .R2 ; d1 / and (iii) .R2 ; d1 /. In
(iv), g 2 B.f; ı/, where f; g 2 C Œ0; 1 and we use the metric d1 .
Definition 1.6. A subset U X of a metric space .X; d / is an open subset
if 8P 2 U , 9 open ball B.P; ı/ U for some ı > 0.
Note that an open subset is just a union of (usually infinitely many) open balls. Here is an
example of an open subset U of the space .R2 ; d2 /:
ı
P
U
From Definition 1.6, we also define that a subset F X is closed if X n F is open.
Example. The closed ball B.P; ı/ ´ fQ 2 X W d.P; Q/ 6 ıg is a closed subset of X .
Indeed, we show its complement is open.
ı0
ı
Q
P
If Q … B.P; ı/, then d.P; Q/ > ı. We take some 0 < ı 0 < d.P; Q/
R 2 B.Q; ı 0 /. Then using the triangle inequality, d.P; R/ > d.P; Q/
d.P; Q/ ı 0 > ı. So B.Q; ı 0 / X n B.P; ı/, as required.
6
ı. Suppose
d.R; Q/ >
1.3
LIMITS AND CONTINUITY
Lemma 1.7. Let .X; d / be a metric space. Then:
(i) X and ; are open subsets of .X; d /;
S
(ii) If, 8i 2 I , Ui are open subsets of .X; d /, then so too is i 2I Ui ;
(iii) If U1 ; U2 X are open, then so too is U1 \ U2 .
Proof. (i) and (ii) are immediate.
For (iii): Let P 2 U1 \ U2 , then 9 open balls B.P; ı1 / U1 and B.P; ı2 / U2 . Take
ı D min fı1 ; ı2 g. Then B.P; ı/ U1 \ U2 .
Definition 1.8. Let P be a point in .X; d /. An open neighbourhood (nbhd)
of P is an open subset N 3 P , for example, the open balls centred at P .
Example. Open neighbourhoods around P 2 R2 with the British rail metric :
If P ¤ O, and 0 < ı < d.P; O/, we have B .P; ı/ D fP g. If P D O, then B .P; ı/
are the open Euclidean discs of radius ı. So U .R2 ; / open means either O … U , then
U can be arbitrary, or O 2 U , then for some ı > 0 the Euclidean disc BEucl .O; ı/ U .
1.3 Limits and continuity
Suppose x1 ; x2 ; : : : is a sequence of points in the metric space .X; d /. We define xn ! x 2 X ,
and say xn converges to limit x, to mean d.xn ; x/ ! 0 as n ! 1. Equivalently, we have
8 > 0, 9N such that xn 2 B.x; / 8n > N . See an example about convergence in .Q; dp / on
page 5.
Example. Consider fn 2 X ´ C Œ0; 1 in the proof of Proposition 1.5 on page 5. fn ! 0
in .X; d1 /, but not so in .X; d2 / or .X; d1 /.
Proposition. xn ! x in .X; d / , 8 open neighbourhoods U 3 x, 9N such
that xn 2 U for all n > N .
Proof. ((): Take U D B.x; / for any given > 0.
()): Given an open set U 3 x, 9 > 0 such that B.x; / U . So 9N such that xn 2
B.x; / U 8n > N .
This means that the convergence of xn may be rephrased solely in terms of open subsets in
.X; d /.
Caveat. Consider X ´ C Œ0; 1, and the space .X; d1 /. Let gn 2 X be as follows:
7
1.3
LIMITS AND CONTINUITY
1
gn
0
1
n
1
Then gn .0/ D 1 8n, but gn ! 0 as n ! 1. Is this not counter-intuitive?
Definition 1.9. (-δ definition of continuity). A function f W .X; 1 / !
.Y; 2 / is
continuous at x 2 X if 8 > 0, 9ı > 0 such that 1 .x 0 ; x/ < ı )
2 .f .x 0 /; f .x// < ;
(ii) uniformly continuous on X if 8 > 0, 9ı > 0 such that 1 .x1 ; x2 /
< ı ) 2 .f .x1 /; f .x2 // < .
(i)
Note that item (i) of the above definition may be rephrased as: f W .X; 1 / ! .Y; 2 / is
continuous at x 2 X iff 8 > 0, 9ı > 0 such that f .B.x; ı// B.f .x/; /, or equivalently,
B.x; ı/ f 1 .B.f .x/; // D fx 0 2 X W f .x 0 / 2 B.f .x/; /g.
Lemma 1.10. If f W .X; 1 / ! .Y; 2 / is continuous and xn ! x in .X; 1 /,
then f .xn / ! f .x/ in .Y; 2 /.
Proof. 8 > 0, 9ı > 0 such that 1 .x 0 ; x/ < ı ) 2 .f .x 0 /; f .x// < . As xn ! x, 9N such
that n > N ) 1 .xn ; x/ < ı. So n > N ) 2 .f .xn /; f .x// < . Therefore f .xn / ! f .x/.
Example. Consider the identity map id W .C Œ0; 1; d1 / ! .C Œ0; 1; d1 /. Since d1 .f; g/
< , supx2Œ0;1 jf .x/ g.x/j < ) d1 .f; g/ < , we see that id is continuous.
We can use the functions fn 2 C Œ0; 1 in the proof of Proposition 1.5 on page 5 to
show that the identity in the other direction, id W .C Œ0; 1; d1 / ! .C Œ0; 1; d1 /, is not
continuous, again by noting that d1 .fn ; 0/ ! 0 but d1 .fn ; 0/ ! 1 as n ! 1.
We wish to express the continuity of a map purely in terms of open sets.
Proposition 1.11. A map f W .X; 1 / ! .Y; 2 / is continuous ,
(i) 8 open subsets U Y , f 1 .U / are open in X; or equivalently
(ii) 8 closed subsets F Y , f 1 .F / are closed in X .
Proof. (i) ((): For given > 0 and x 2 X, take U D B.f .x/; /. Then f 1 .U / is open )
9ı > 0 such that B.x; ı/ f 1 .U /. This means 1 .x 0 ; x/ < ı ) 2 .f .x 0 /; f .x// < .
()): For an open subset U 2 Y and any x 2 f 1 .U /, we can choose an open ball
B.f .x/; / U . Since f is continuous at x, 9ı > 0 such that B.x; ı/ f 1 .B.f .x/; // f 1 .U /. This is true for all such x, therefore f 1 .U / is open in X .
8
1.4
COMPLETENESS
(ii) ()): F is closed in Y , Y n F is open in Y . By (i), this implies f
equal to X n f 1 .F /, is open in X , f 1 .F / is closed in X.
1
.Y n F /, which is
((): Suppose U is open in Y . Then Y n U is closed in Y so by the hypothesis, f
X n f 1 .U / is closed in X , f 1 .U / is open in X . By (i), f is continuous.
1
.Y n U / D
1.4 Completeness
A metric space .X; d / is called complete if, for all sequences of the form x1 ; x2 ; : : : in X
satisfying “8 > 0, 9N such that d.xm ; xn / < 8m; n > N ”, we have xn ! x for some x 2 X .
.R; d1 / is complete. This is known as Cauchy’s principle of convergence. However, .Q; d1 /
and ..0; 1/ R; dEucl / are both not complete.
Example. Let X ´ C Œ0; 1. We show that .X; d1 / is not complete.
For n > 1, take fn 2 X as follows:
1
fn
0
1
2
1
2
C
1
n
1
Then d1 .fm ; fn / 6 N1 for m; n > N , so fn form a Cauchy sequence. Suppose
R1
R1
fn ! f 2 X as n ! 1, i.e. 0 jfn f j ! 0. By the triangle inequality, 0 jfn
R 1=2
R1
R 1=2
R1
f j > 0 .jf
1j jfn 1j/ C 1=2 .jf j jfn j/ ! 0 jf
1j C 1=2 jf j, giving
R 1=2
R1
1j D 1=2 jf j D 0.
0 jf
Assuming f is continuous, this means
(
f .x/ D
1 if x < 12 ;
0 if x > 12 ;
which is discontinuous at x D 12 . Contradiction.
§2
T OPOLOGICAL SPACES
2.1 Introduction
Consider the open subsets of a metric space with properties as described in Lemma 1.7 on
page 7. These properties may be abstracted out for a definition of a topological space:
Definition 2.1. A topological space .X; / consists of a set X and a set (the
“topology”) of subsets of X (hence P.X/, the power set of X), where
by definition we call the elements of the “open” subsets, satisfying:
9
2.1
INTRODUCTION
(i) X; ; 2 ;
S
(ii) If Ui 2 8i 2 I , then i 2I Ui 2 ;
(iii) If U1 ; U2 2 , then U1 \ U2 2 .
By induction, we also get the closure property (iii) for finite intersections.
A subset Y X in a topological space .X; / is called closed if X n Y is open. Therefore we
can describe a topology on a set X by specifying the closed sets in X which satisfy
(i) X; ; are closed;
T
(ii) If Fi are closed 8i 2 I , then so too is i 2I Fi ;
(iii) If F1 ; F2 are closed, then so too is F1 [ F2 .
This description of a topology is sometimes more natural: Consider the non-metric topologies in
examples (ii) and (iii) below.
Lemma 1.7 on page 7 implies that every metric space .X; d / gives rise to a topology. This is
called the metric topology.
Two metrics 1 ; 2 on X are said to be topologically equivalent if their associated topologies
are the same.
Exercise. Show that 1 ; 2 are Lipschitz equivalent ) 1 ; 2 are topologically equivalent.
Example. The discrete metric on a set X gives rise to the discrete topology in which
every subset in X is open, i.e. D P.X/.
Examples of non-metric topologies
(i)
Let X be a set with at least two elements, and ´ fX; ;g, the indiscrete topology.
(ii) Let X be any infinite set, and ´ f;g [ fY X W X n Y is finiteg, the co-finite
topology. If X D R or C then this is known as the Zariski topology, where open
sets are “complements of zeros of polynomials”. The Zariski topologies on Rn or Cn
are very important in algebraic geometry.
(iii) Let X be any uncountable set e.g. R or C, and ´ f;g [ fY X W X n Y is
countableg, the co-countable topology.
˚
(iv) For the set X D fa; bg, there are exactly 4 distinct
topologies:
´
;; fag; fbg; fa;
˚
bg i.e.
´ ;; fa; bg i.e. the indiscrete topology,
˚ the discrete/ metric topology,
˚
´ ;; fag; fa; bg , or ´ ;; fbg; fa; bg .
Example. The half-open interval topology on R consists of arbitrary unions of halfopen intervals Œa; b/ where a < b, a; b 2 R. Clearly R; ; 2 and is closed under unions.
Suppose U1 ; U2 2 . We show that 8P 2 U1 \U2 , 9 Œa; b/ such that P 2 Œa; b/ U1 \U2
and hence U1 \ U2 2 .
Since P 2 U1 and P 2 U2 , we have P 2 Œa1 ; b1 / U1 and P 2 Œa2 ; b2 / U2
for some a1 < b1 , a2 < b2 . Let a D max fa1 ; a2 g and b D min fb1 ; b2 g. Then
P 2 Œa; b/ U1 \ U2 , as required.
10
2.1
INTRODUCTION
More definitions
Definition 2.2. Let .X; 1 / and .Y; 2 / be topological spaces. Let P 2 .X; 1 /.
An open neighbourhood (nbhd) of P is an open set U X with
P 2 U . (cf. Definition 1.8 on page 7.)
(ii) A sequence of points xn converges to limit x if, for any open neighbourhood U 3 x, 9N such that xn 2 U 8n > N . (cf. Section 1.3 on
page 7.)
(iii) A map f W .X; 1 / ! .Y; 2 / is continuous if, for any open set
U Y , f 1 .U / is open in X . (cf. Proposition 1.11(i) on page 8.)
(i)
The proof of Proposition 1.11(ii) on page 8 shows that f is continuous , 8 closed set F Y ,
f 1 .F / is closed in X.
Example. The identity map id W .R; Eucl / ! .R; co-finite topology/ is continuous,
because closed sets in the co-finite topology, namely, the finite sets and R, are closed in
the Euclidean topology.
The identity map id W .R; Eucl / ! .R; co-countable topology/ is not continuous,
because Q R is closed in the co-countable topology, but not so in the Euclidean
topology.
Definition 2.3. A map f W .X; 1 / ! .Y; 2 / is a homeomorphism if:
(i) f is bijective;
(ii) Both f and f
1
are continuous.
In this case, the open subsets of X correspond precisely to the open subsets of Y under the
bijection f . This can be used to define an equivalence relation between topological spaces.
A property on topological spaces is called a topological property if .X; 1 / has the property,
and .X; 1 / is homeomorphic to .Y; 2 / implies that .Y; 2 / has the same property.
Example. Consider the topological spaces .R; Eucl / and . 1; 1/; Eucl . Let f W R !
. 1; 1/, defined by f .x/ D x=.1 C jxj/. f is bijective with inverse g W . 1; 1/ ! R
given by g.y/ D y=.1 jyj/. Both f and g are continuous, and so f and g are
homeomorphisms.
Note that .R; dEucl / is complete, whilst . 1; 1/; dEucl is not. Therefore this example
shows that ‘completeness’ is a property of the metric on a metric space, and not just a
topological property.
However, it also shows that using the homeomorphism, we can construct a complete
metric on . 1; 1/ coming from dEucl on R which is topologically equivalent to dEucl on
. 1; 1/.
Definition 2.4. A topological space .X; / is called Hausdorff if 8P; Q 2 X
with P ¤ Q, 9 disjoint open sets U 3 P and V 3 Q, i.e. we can separate
points by open sets. This is a topological property.
11
2.1
INTRODUCTION
U
V
Q
P
Example. R with the indiscrete, co-finite or co-countable topology is not Hausdorff,
because any two non-empty open sets intersect non-trivially (see the Examples (i), (ii)
and (iii) on page 10). But any metric space is clearly Hausdorff. So these topologies are
non-metric.
Example. The half-open interval topology on R is Hausdorff. If a; b 2 R with a < b,
then take Œa; b/ 3 a, Œb; b C 1/ 3 b will do. Example Sheet 2, Question 18 shows that this
topology is non-metric.
Definition 2.5. In a topological space X, for any A X , we say that x0 2 X
is an accumulation point or limit point of A if any open neighbourhood U of
x0 satisfies U \ A ¤ ;.
Lemma 2.6. A X is closed , A contains all of its accumulation points, i.e.
if x0 2 X is an accumulation point of A then x0 2 A.
Proof. ()): Suppose A is closed and x0 2 X n A. Then take U ´ X n A, an open
neighbourhood of x0 . Now U \ A D ;, so x0 is not an accumulation point.
((): Suppose A is not closed, then X n A is not open. 9x0 2 X n A such that no open
neighbourhood U of x0 is contained in X n A, i.e. any open neighbourhood U of x0 satisfies
U \ A ¤ ;. So x0 is an accumulation point of A but not in A.
Remark. Suppose we have a convergent sequence xn ! x 2 X with xn 2 A 8n. Then for
any open neighbourhood U of x, 9N such that xn 2 U 8n > N . So x is an accumulation
point of A. Therefore if A is closed, we must have x 2 A.
A topological space .X; / has a countable base of open neighbourhoods (or is first countable)
if, 8P 2 X, 9 open neighbourhoods N1 N2 N3 of P with the property that, 8 open
neighbourhood U of P , 9m such that Ni U 8i > m.
Example. Any metric space is first countable: Given P 2 X , take the open balls B P; n1
for n 2 N.
Lemma 2.7. If .X; / is first countable, and A X has the property that “8
convergent sequences xn ! x 2 X with xn 2 A 8n, the limit x is in A”, then
A is closed.
Proof. By Lemma 2.6, it suffices to prove that any accumulation point x of A is the limit of
some sequence xn with xn 2 A 8n. Let N1 N2 N3 be a base of open neighbourhoods
for x. Then 8i , since x is an accumulation point, 9xi 2 A \ Ni , giving the sequence .xi /. 8
open neighbourhood U of x, by first countability 9m such that Ni U 8i > m. So xi 2 U
8i > m, i.e. xn ! x.
Remark. Example Sheet 1, Question 19 (revised version on the DPMMS website) shows
that we do need the first countability condition here.
12
2.2
INTERIORS AND CLOSURES
2.2 Interiors and closures
Given any A X , we define the interior Int .A/ or A0 of A to be the union of all open subsets
contained in A. Int .A/ is the largest open subset contained in A: If U is open and U A, then
U Int .A/.
Define the closure Cl .A/ or A of A to be the intersection of all closed sets containing A.
Cl .A/ is the smallest closed set containing A: If F is closed and A F , then Cl .A/ F .
Example. Consider R with the Euclidean topology. Int .Q/ D ; (because any nonempty open subset contains irrationals) and Cl .Q/ D R. Also, Int .Œ0; 1/ D .0; 1/ and
Cl ..0; 1// D Œ0; 1.
For any set, we have Int .A/ A Cl .A/. The boundary or frontier of A is defined to be
@A ´ Cl .A/ n Int .A/.
A set A X is called dense if Cl .A/ D X.
If A B are subsets of X , then Int .A/ Int .B/ and Cl .A/ Cl .B/.
Proposition 2.8. Let A X. Then:
(i) Int .Cl .Int .Cl .A//// D Int .Cl .A//;
(ii) Cl .Int .Cl .Int .A//// D Cl .Int .A//.
Proof. (i): Since Int .Cl .A// is open and Int .Cl .A// Cl .Int .Cl .A///, taking interiors of
both sides we have Int .Cl .A// Int .Cl .Int .Cl .A////. Since Cl .A/ is closed and Int .Cl .A// Cl .A/, taking closures then interiors of both sides we have Int .Cl .Int .Cl .A//// Int .Cl .A//.
(ii): Similar argument. See Example Sheet 1, Question 11.
This shows that if we start from an arbitrary set A X and take successive interiors and closures, we may obtain at most 7 distinct sets, namely, A, Int .A/, Cl .A/, Cl .Int .A//, Int .Cl .A//,
Int .Cl .Int .A/// and Cl .Int .Cl .A///. There is such an A where all 7 of these are distinct: See
Example Sheet 1, Question 11.
2.3 Base of open subsets for a topology
Given a topological space .X; /, we say that a collection B ´ fUi gi 2I of “basic” open sets
form a base or basis for the topology if any open set is the union of elements of B.
Question: When does an arbitrary collection B ´ fUi gi 2I of subsets of a set X form the base
for some topology on X ? Answer: If, 8i; j , Ui \ Uj is the union of some Uk ’s in B. If so, we
can define a topology by specifying that an open set is just a union of some Ui ’s in B (and also
include X , if necessary).
A topological space .X; / is called second countable if it has a countable base of open sets.
.X; / is second countable ) .X; / is first countable, because 8P 2 X and open neighbourhood
U of P , U is the union of some basic open sets and so 9Ui U with P 2 Ui 8i 2 N.
13
2.4
S U B S PA C E S , Q U O T I E N T S A N D P R O D U C T S
2.4 Subspaces, quotients and products
Let .X; / be a topological space, and Y X . The subspace topology on Y is jY ´
fU \ Y W U 2 g. Consider the inclusion map i W Y ,! X . It is continuous (because
i 1 .U / D U \ Y ), and the subspace topology is the smallest topology on which i is continuous.
Proposition 2.9.
Let B be a base for the topology . Then BjY ´ fU \ Y W U 2 Bg
is a base for the subspace topology;
(ii) Let .X; / be a metric space and 1 be the restriction of the metric
to Y . Then the subspace topology on Y induced from the -metric
topology on X is the same as the 1 -metric topology on Y .
(i)
Proof. (i): Any open set U X is U D
S
˛
U˛ where U˛ 2 B, so U \ Y D
S
˛ .U˛
\ Y /.
(ii): A base for the metric topology on X is given by open balls B .x; ı/ for a fixed x 2 X .
8y 2 B .x; ı/ \ Y , 9ı 0 > 0 such that B .y; ı 0 / B .x; ı/, then B1 .y; ı 0 / D B .y; ı 0 / \ Y B .x; ı/ \ Y . Therefore B .x; ı/ \ Y , a base for the subspace topology, is the union of open
1 -balls.
Example. .0; 1 R with the Euclidean topology has a base for the subspace topology
consisting of open intervals .a; b/ and half-open intervals .a; 1, where 0 < a < b < 1.
Let .X; / be a topological space and be an equivalence relation on X . Denote the quotient
set by Y ´ X= and the quotient map by q W X ! Y where x 7! Œx. The quotient topology
on Y is then given by fU Y W q 1 .U / 2 g, i.e. the subsets U of the quotient set Y , for which
the union of the equivalence classes in X corresponding to points of U is an open subset of X.
Remark. q is continuous, and the quotient topology is the largest topology on Y for
which this is so. If f W X ! Z is any continuous map between topological spaces
such that x y ) f .x/ D f .y/, then there is a unique factorisation and f , defined
1
by f .Œx/ D f .x/, is continuous. (q 1 .f .U // D f 1 .U / is open in X ) f is
continuous.)
X
q
f
X= 9Šf
Z
Examples.
(i)
Define on R by x y iff x y 2 Z. The map W R= ! T ´ fz 2 C W jzj D 1g,
where Œx 7! e 2 ix , is well-defined and a homeomorphism. (See Example Sheet 1,
Question 15).
(ii) Define the 2-D torus T to be R2 = where .x1 ; y1 / .x2 ; y2 / iff x1 x2 2 Z and
y1 y2 2 Z. The topology in fact comes from a metric on T (See Example Sheet 1,
Question 18) and hence well-behaved.
14
2.4
S U B S PA C E S , Q U O T I E N T S A N D P R O D U C T S
In general, one can get rather nasty (non-Hausdorff, for instance) topologies from an arbitrary
equivalence relation.
If A X we can define on X by x y iff x D y or x; y 2 A. The quotient space is
sometimes written as X=A, in which A is scrunched down to a point. Usually closed A is taken.
Example. Let D to be the closed unit disc in C, with boundary C , the unit circle. Then
D=C is homeomorphic to S 2 , the 2-sphere. (See Example Sheet 2, Question 13.)
Proposition 2.10. Suppose .X; / is Hausdorff and A X is closed, and
8x 2 X n A, 9 open sets U; V with U \ V D ;, A U , x 2 V , then X=A is
Hausdorff.
Proof. Given any two points x ¤ y in X=A, either one of the following holds:
(i)
Neither x or y correspond to A: then 9Šx; y 2 X corresponding to x; y. Now
9Ux A and Vx 3 x such that Ux \ Vx D ;. 9Uy A and Vy 3 y such
that Uy \ Vy D ;. Since X is Hausdorff, wlog we assume Vx \ Vy D ;. The
corresponding open sets q.Vx /; q.Vy / in X=A separate x and y.
(ii) x D q.x/ where x 2 X n A and y corresponds to A. Then 9 open sets U A and
V 3 x such that U \ V D ;. The corresponding open sets q.U /; q.V / 2 X=A
separate x and y.
Given topological spaces .X; / and .Y; /, we define the product topology on X Y
by W X Y is open iff 8.x; y/ 2 W , 9 open sets U X and V Y such that x 2 U ,
y 2 V and U V W . A base of open sets for .X Y; / consists of sets U V , where
U is open in X and V is open in Y . Note that .U1 V1 / \ .U2 V2 / D .U1 \ U2 / .V1 \ V2 /,
so this does define a base for the topology.
Similarly if .Xi ; i /niD1 are topological spaces, the product topology on X1 Xn has a
base of open sets of the form U1 Un , where Ui is open in Xi .
Example. Consider R with the Euclidean topology. The product topology for R R D R2
is just the metric topology, where open rectangles I1 I2 , with I1 and I2 being open
intervals in R, form its base and also a base for the Euclidean topology on R2 .
Y
)
W
.x; y/
(
V
U V
(
)
U
15
X
CONNECTEDNESS
Lemma 2.11. Given topological spaces .X1 ; 1 / and .X2 ; 2 /, the projection
maps i W .X1 X2 ; 1 2 / ! .Xi ; i / are continuous. Given a topological
space .Y; / and continuous maps fi W Y ! Xi , there is a unique factorisation,
and f (in the diagram) is continuous.
f1
Y
9Šf
X1
X1 X2
f2
1
2
X2
Proof. For the first part, use 1 1 .U / D U X2 and 2 1 .V / D X1 V . For the second part,
we have f .y/ D .f1 .y/; f2 .y//. 8 basic open set U V X1 X2 , where U is open in X1 and
V is open in X2 , f 1 .U V / D f1 1 .U / \ f2 1 .V / is open in Y . Since any any open set W in
X1 X2 is the union of basic open sets, f 1 .W / is the union of their inverses, and so f 1 .W /
is open in Y . Therefore f is continuous.
§3
C ONNECTEDNESS
3.1 Various results
Definition 3.1. A topological space X is disconnected if 9 non-empty open
subsets U; V such that U \ V D ; and U [ V D X . We say U; V disconnect
X . Otherwise X is connected.
From this definition, we have X is connected iff 8 open sets U; V where U \ V D ; and
U [ V D X , either U D ; (whence V D X) or V D ; (whence U D X ). Connectedness is a
topological property.
If Y X , then Y is disconnected in the subspace topology iff 9 open sets U; V X such that
U \ Y ¤ ;, V \ Y ¤ ;, U \ V \ Y D ; and Y U [ V . We say that U; V disconnect Y .
Proposition 3.2. Let X be a topological space. Then the following are equivalent:
(i) X is connected;
(ii) The only subsets of X which are both open and closed are ;; X;
(iii) Every continuous function f W X ! Z is constant.
Proof. (i) , (ii): Trivial. For U X , consider U [ .X n U / D X .
Not (iii) ) Not (i): Suppose 9 non-constant f W X ! Z which is continuous, then 9m; n 2
f .X / with m < n. f 1 .fk W k 6 mg/ and f 1 .fk W k > mg/ are non-empty, open sets
disconnecting X .
16
3.1
VA R I O U S R E S U LT S
Not (i) ) Not (iii): 9 non-empty disjoint open sets U; V such that U [V D X. Let f W X ! Z
where f .x/ D 0 if x 2 U , f .x/ D 1 if x 2 V . This is continuous but non-constant. (It is locally
constant, however.)
Proposition 3.3. Continuous images of a connected space are connected.
Proof. If f W X ! Y is a continuous surjective map between topological spaces, and if U; V
disconnect Y , then f 1 .U /; f 1 .V / disconnect X . So X is connected ) Y is connected.
Connectedness in R
A subset I R is called an interval if given x; z 2 I where x 6 z, we have y 2 I 8y
satisfying x 6 y 6 z. There are the cases inf I D a 2 R and a 2 I , or a … I , or inf I D 1.
Similarly we have the cases sup I D b 2 R and b 2 I , or b … I , or sup I D 1. So any interval
I takes the form Œa; b, Œa; b/, .a; b, .a; b/, Œa; 1/, .a; 1/, . 1; b, . 1; b/ or . 1; 1/.
Theorem 3.4. A subset of R is connected , it is an interval.
Proof. ()): If X R is not an interval, 9x; y; z where x < y < z, and x; z 2 X but y … X .
Then . 1; y/ and .y; 1/ disconnect X .
((): Let I be an interval and U; V be open subsets of R disconnecting I . 9u 2 U \ I and
v 2 V \ I where, wlog, u < v. Since I is an interval, Œu; v I . Let s D sup .Œu; v \ U /.
If s 2 U , then s ¤ v so s < v and, since U is open, 9ı > 0 such that .s
9s 0 2 Œu; v \ U with s 0 > s. This contradicts s being an upper bound.
If s 2 V , then 9ı > 0 such that .s ı; s C ı/ V . In particular, .s
Œu; v \ U Œu; s ı. This contradicts s being the least upper bound.
ı; s C ı/ U . So
ı; s C ı/ \ U D ;, so
Corollary 3.5. (Intermediate value theorem). Let a < b and f W Œa; b ! R
be continuous. If y 2 Œf .a/; f .b/ (or Œf .b/; f .a/) then 9x 2 Œa; b such that
f .x/ D y.
Proof. Suppose not, then f 1 . 1; y/ and f 1 .y; 1/ disconnect Œa; b.
Connected subsets and subspaces
Proposition 3.6. Given connected subspaces fY˛ gS
˛2A of a topological space
X, where Y˛ \ Yˇ ¤ ; 8˛; ˇ 2 A, the union Y D ˛ Y˛ is connected.
Proof. Using Proposition 3.2(iii), it suffices to prove that any continuous function f W Y ! Z
is constant. Using the same proposition, f˛ ´ f jY˛ is constant, equal to n˛ , say, on Y˛ 8˛ 2 A.
Since 8ˇ ¤ ˛, 9z 2 Y˛ \ Yˇ , we have n˛ D f .z/ D nˇ . Therefore f is constant on Y .
A connected component of a topological space X is a maximal connected subset Y X , i.e.
if Z X is connected and Z Y then Z D Y .
S
Each point x 2 X is contained in a unique connected component of X , namely, fZ X W Z
is connected and x 2 Zg. The fact that this is connected is a result of Proposition 3.6.
17
3.1
VA R I O U S R E S U LT S
˚
Example. Let X ´ f0g [ n1 W n D 1;˚2;: : : R, with the subspace topology. The
connected component containing n1 is n1 , both open and closed in X . The connected
component containing 0 is f0g which is closed but not open in X .
Proposition 3.7. If Y is a connected subset of a topological space X, then the
closure Y is connected.
Proof. Suppose f W Y ! Z is continuous. Proposition 3.2(iii) on page 16 ) f jY is constant,
equal to m, say. Given x 2 Y , let n ´ f .x/, then f 1 .n/ is an open neighbourhood of x in Y , i.e.
of the form U \ Y where U is open in X . Since U \ Y ¤ ;, (considering Y D X n Int .X n Y /)
it contains a point z 2 Y at which f .z/ D m, so n D m, i.e. f is constant on Y .
This shows that connected components of a topological space must be closed (because they are
maximal connected subsets), but are not necessarily open, as the above example shows.
We call a topological space X totally disconnected if its connected components are just the
single points, or equivalently, its only connected subsets are single points. Any discrete topological
space is totally disconnected, as is the above example.
Lemma. If X is a topological space, and 8x; y 2 X where x ¤ y, X may
be disconnected by U; V X where x 2 U and y 2 V , then X is totally
disconnected.
Proof. For any Y X with points x; y where x ¤ y, take U; V as given. Then U; V
disconnect Y .
The set of irrationals R n Q is totally disconnected: Use the above lemma and the fact that
between any two distinct irrationals there is a rational.
Cantor set
1 2 1 2
;
Consider starting with I0 ´ Œ0; 1
and
removing
to
get
I
´
0; 3 [ 3 ; 1 . Now
1
1
2 3 3
remove
Tthe middle third from both 0; 3 and 3 ; 1 to get I2 , and so on. The Cantor set is
C ´ n>0 In .
0
I0
Œ
I1
Œ
I2
Œ
1
9
2
9
1
3
4
9
5
9
2
3
7
9
8
9
1


Œ

Œ

Œ


Œ

We can understand this in terms of ternary (i.e. base 3) expansions. I1 consists of numbers
with ternary expansion 0:a1 a2 a3 : : : where a1 D 0 or 2. (Wlog we impose 13 D 0:022 : : : .) I2
consists of numbers where a1 D 0 or 2, and a2 D 0 or 2, and so on for I3 etc. So C consists of
numbers with a ternary expansion where each ai D 0 or 2.
Suppose now we are given two distinct points x; y in C . For some n, the ternary expansions
will differ first in the n-th place. C In , and In consists of 2n disjoint closed intervals, one of
which contains x and another contains y.
18
3.2
P AT H - C O N N E C T E D N E S S
So we can disconnect In by open U; V Œ0; 1 where x 2 U \ In and y 2 V \ In . This
implies U; V disconnect C where x 2 U \ C and y 2 U \ C . By the previous lemma, C is
totally disconnected. Also, both C and its complement are uncountable: Consider the ternary
expansions.
3.2 Path-connectedness
Let X be a topological space and x; y 2 X. A continuous path from x to y is a continuous
function W Œa; b ! X such that .a/ D x, .b/ D y. (You may take a D 0, b D 1 if you
prefer.) We say X is path-connected if 8x; y 2 X , 9 a path from x to y. This is a topological
property.
Proposition 3.3∗ . Continuous images of a path-connected space are pathconnected. (cf. Proposition 3.3 on page 17.)
Proof. Suppose f W X ! Y is a continuous surjection. Given y1 ; y2 2 Y , pick xi 2 f 1 .yi /
for i D 1; 2, and a path W Œa; b ! X from x1 to x2 . Then f B W Œa; b ! Y is a path from y1
to y2 .
Proposition 3.8. Path-connected ) Connected.
Proof. Let X be a path-connected topological space, and suppose U; V X disconnect X .
Choose some u 2 U , v 2 V . Then 9 a continuous function W Œa; b ! X with .a/ D u and
.b/ D v. Now 1 .U /, 1 .V / Œa; b disconnect Œa; b, contradiction.
But connected ; path-connected:
Example. In R2 , let I ´
W 0 < y 6 1g. For
˚ f.x;0/ W 0 < x 6 1g and J ´ f.0; y/ S
n D 1; 2; : : : , let Ln ´ n1 ; y W 0 6 y 6 1 . Set X D I [ J [
n>1 Ln , with the
subspace topology.
1
J
0
1
4
1
3
1
2
1
I
X is not path-connected: For any continuous path .t/ D 1 .t/; 2 .t/ from .1; 0/ to
.0; 1/, 9s such that 1 .s/ D 0 and 1 .t/ > 0 8t < s, so we must also have 2 .s/ D 0.
This means passes through .0; 0/ … X .
X is connected: S
Supposef W X ! Z is a continuous function. f is constant
on J
1 1
1
and on Y ´ I [
L
.
However,
the
points
;
2
Y
have
limit
0;
2 J as
n>1 n
n 2
2
n ! 1, hence continuity of f ) the two constants agree, i.e. f is constant on X .
19
3.3
P R O D U C T S O F C O N N E C T E D S PA C E S
Given a topological space X, we can define an equivalence relation on X by x y iff 9 a
path from x to y in X. Reflexivity of is trivial. Symmetry of : If W Œa; b ! X is a path
from x to y, then .t/ D . t/ gives a path W Œ b; a ! X from y to x. Transitivity of
: If W Œa; b ! X and W Œc; d  ! X are paths from x to y and y to z respectively, then
W Œa; b C d c ! X where
(
.t/
if t 2 Œa; b;
.t/ ´
.t C c b/ if t 2 Œb; b C d c;
is a path from x to z. The equivalence classes under are called the path-connected components
of X .
Theorem 3.9. Let X be an open subset of the Euclidean space Rn , then X is
connected , X is path-connected.
Proof. ((): Proved as Proposition 3.8. ()): Let x 2 X and U be the equivalence class of x
under defined as above.
U is open in X : Let y 2 U , whence x y. Since X is open, 9ı > 0 such that B.y; ı/ X .
Then 8z 2 B.y; ı/, we have y z (take the straight line segment), so transitivity of ) x z,
i.e. B.y; ı/ U .
Similarly, X n U is open in X. Now X is connected and U ¤ ;, so we have X n U D ; ,
U D X , x y 8y 2 X .
3.3 Products of connected spaces
Let X; Y be topological spaces. Consider X Y with the product topology.
Proposition 3.10. If X; Y are path-connected, then so too is X Y .
Proof. Let .x1 ; y2 / and .x2 ; y2 / be in X Y . 9 paths 1 W Œ0; 1 ! X and 2 W Œ0; 1 ! Y
from x1 to x2 and y1 to y2 , respectively. Define W Œ0; 1 ! X Y by .t/ ´ 1 .t/; 2 .t/ .
Any base for the topology on X Y consists of open sets U V where U is open in X and V is
open in Y . Now 1 .U V / D 1 1 .U / \ 2 1 .V / is open. So is continuous and gives a path
from .x1 ; y1 / to .x2 ; y2 /.
Proposition 3.11. If X; Y are connected, then so too is X Y .
First we make some general comments about the product topology X Y . 8y 2 Y , X fyg
with the subspace topology is homeomorphic to X via the projection map 1 W X fyg ! X .
Similarly, 8x 2 X, fxg Y is homeomorphic to Y . So X , Y are connected ) X fyg and
fxg Y are connected.
Proof of 3.11. Let f W X Y ! Z be continuous. By connectedness, f is constant on each
of fxg Y and X fyg. Take two points .x1 ; y1 / and .x2 ; y2 / in X Y , then f .x1 ; y1 / D
f .x1 ; y2 / D f .x2 ; y2 / (See the following diagram), hence f constant on X Y , and X Y is
connected.
20
C O M PA C T N E S S
Y
y2
.x2 ; y2 /
y1
.x1 ; y1 /
0
x1
x2
X
A similar argument proves Proposition 3.10: 9 paths from .x1 ; y2 / to .x1 ; y2 / and from .x1 ; y2 /
to .x2 ; y2 / in X Y .
§4
C OMPACTNESS
4.1 Various results
Definition 4.1. Let X be a topological space and Y X . AnSopen cover of
Y is a collection fU W 2 g of open subsets such that Y 2 U .
Such an open cover of Y provides a base of open sets fU \ Y W 2 g for Y with the
subspace topology, and conversely. A subcover of an open cover U D fU W 2 g of Y is a
subcollection V U which is still an open cover of Y .
Example. The intervals In D . n; n/ where n D 1; 2; : : : form an open cover of R, and
In2 is a proper sub-cover. The intervals Jn D .n 1; n C 1/ where n 2 Z form an open
cover of R with no proper subcover.
Definition 4.2. A topological space X is compact if every open cover has a
finite subcover.
By the above example, R is not compact. Any finite topological space is compact, as is any set
with the indiscrete or co-finite topology.
By the way we defined it, compactness is a topological property.
Lemma 4.3. Let X be a topological space. Then Y X with the subspace
topology is compact , every open cover fU g of Y has a finite subcover.
S
Proof. ()): Let fU W 2 g be an open cover of Y , then Y D 2 fU \ Y g, and the
S
U \ Y ’s are open in Y . Since Y is compact, 91 ; : : : ; n 2 such that Y D niD1 Ui \ Y )
fUi W i D 1; : : : ; ng covers Y .
S
((): Suppose Y D 2 V where the V ’s are open in Y . Write V D U \ Y where the
Sn
U ’s are open
in
X
and
form
an
open
cover
of
Y
.
So
9
;
:
:
:
;
2
such
that
Y
1
n
i D1 Ui ,
S
hence Y D niD1 Vi .
The open
interval .0; 1/ R is not compact: Consider the open cover consisting of intervals
1
for n D 3; 4; : : : .
n
1
;1
n
21
4.1
VA R I O U S R E S U LT S
Theorem 4.4. (Heine-Borel). The closed interval Œa; b R is compact.
S
Proof. Let Œa; b 2 U , where the U ’s are open in R. Let K ´ fx 2 Œa; b W Œa; x is
covered by finitely many U ’sg. K ¤ ; since a 2 K. K is bounded above by b. Let r ´ sup K,
then r 2 Œa; b, so r 2 U1 for some 1 . Since U1 is open, 9ı > 0 such that Œr ı; r C ı U1 .
¼/
.
U 1
a
r
ı r r Cı
b
By definition of sup K, 9c 2 Œr ı; r such that Œa; c is covered by finitely many of the U ’s.
Including U1 , we have Œa; r C ı \ Œa; b is covered by finitely many of the U ’s. This contradicts
r being an upper bound unless r D b, in which case the above argument says that Œa; b is covered
by finitely many of the U ’s. So Œa; b is compact.
Proposition 4.5. Continuous images of a compact set are compact.
Proof. Suppose f W X !S
Y is a continuous map between topological spaces,
and K X is
S
compact. Suppose f .K/ 2 U , where U is open in Y . Then K 2 f 1 .U /, and
since f isS
continuous, each f 1 .U / isSopen in X . Since K is compact, 91 ; : : : ; n 2 such
that K niD1 f 1 .Ui /. So f .K/ niD1 Ui .
Proposition 4.6. Closed subsets of a compact topological space are compact.
Proof. Let X be a compact topological space,
S and K X be closed. If K D ; then
this is trivial, so assume not. Suppose K 2 U , where the U ’s are open in X . Then
S
X D .X n K/ [
, where X n K is also open. Since X is compact, 91 ; : : : ; n 2
2 U S
Sn
n
such that X D .X n K/ [
i D1 Ui . So K i D1 Ui .
Proposition 4.7. Compact subsets of a Hausdorff space are closed.
Proof. Suppose X is a Hausdorff space, and K X is compact. If K D X then this is trivial,
so assume not. We show that X n K is open.
Let x 2 X n K. Since X is Hausdorff, 8y 2 K, 9 disjoint open sets Uy 3 x and Vy 3 y.
Now fV
of K. Since K is compact, 9y1 ; : : : ; yn 2 K such that
Syn W y 2 Kg is an open
Tcover
n
K i D1 Vyi . Then U ´ i D1 Uyi is an open neighbourhood of x with U \ K D ;, so
U X n K, as required.
Corollary 4.8. A subset X R is compact , X is closed and bounded.
Proof. ()): Since R is Hausdorff, Proposition 4.7 ) X is closed. Suppose X is not bounded,
then the intervals . n; n/ where n D 1; 2; : : : form an open cover of X with no finite subcover.
Contradiction.
((): 9M > 0 such that X Œ M; M . Since R n X is open, we have .R n X/ \ Œ M; M 
is open in Œ M; M , i.e. X is closed in Œ M; M . By Heine-Borel (Theorem 4.4), Œ M; M  is
compact. By Proposition 4.6, X is compact.
22
4.1
VA R I O U S R E S U LT S
Together with Theorem 3.4 on page 17, this shows that the only connected compact subsets of
R are the closed intervals. See also the remark after Theorem 4.11 on page 24.
T
Example. Consider the Cantor set C D n>0 In as defined on page 18. Each In is the
disjoint union of 2n closed intervals, so In is closed. So C is closed and bounded ) C is
compact.
Corollary 4.9. Let f W X ! Y be a continuous bijection from a compact
space X to a Hausdorff space Y . Then f is a homeomorphism.
Proof. Write g ´ f 1 . Let F X be closed. X is compact, so Proposition 4.6 ) F is
compact. f is continuous, so Proposition 4.5 ) g 1 .F / D f .F / is compact. Y is Hausdorff, so
Proposition 4.7 ) g 1 .F / is closed in Y . So g is continuous.
This result is particularly useful in identifying quotient spaces. E.g. If we define on R by
x y iff x y 2 Z, and T ´ fz 2 C W jzj D 1g (the unit circle with the subspace topology
coming from C), then it is clear that the map f W R ! T, given by x 7! e 2 ix , is continuous. It
induces a bijection f W R= ! T, which, by definition of quotient topology, is also continuous.
(See the remark about quotient topologies on page 14.)
The restricted quotient map Œ0; 1 ! R= is a continuous surjection. Œ0; 1 is compact, so
Proposition 4.5 ) R= is compact. T is Hausdorff, so Corollary 4.9 ) f is a homeomorphism.
Similarly, we can show that the 2-D torus R2 =Z2 , defined on page 14, is homeomorphic both
to the product space S 1 S 1 (where S 1 is the unit circle) and to the embedded torus X R3
consisting of points .2 C cos / cos ; .2 C cos / sin ; sin , where ; 2 Œ0; 2/.
(i)
(ii)
Figure 2: (i) The 2-D torus R2 =Z2 ; (ii) The embedded torus X R3 .
In Analysis, you learnt that continuous real-valued functions on Œa; b R are bounded and
attain their bounds. What is being used here is the compactness of Œa; b. This statement is still
true for continuous real-valued functions on the Cantor set, for instance.
Proposition 4.10. Continuous real-valued functions on a compact space X is
bounded and attain their bounds.
Proof. Let X be a compact space, and f W X ! R be continuous. Proposition 4.5 ) f .X/
is compact. Corollary 4.8 ) f .X/ is closed and bounded. Now f .X/ is closed , f .X/
contains all its accumulation points. sup ff .X/g; inf ff .X/g (exist because f .X/ is bounded)
are accumulation points for f .X/, so sup ff .X/g; inf ff .X/g 2 f .X/, hence the result.
23
4.1
VA R I O U S R E S U LT S
Theorem 4.11. The product of two compact spaces is compact.
S
Proof. Let X; Y be compact spaces and X Y D 2 U , where the U ’s are open in X Y .
Each U is the union
S of open sets of the form V W , where V is open in X and W is open in
Y . So X Y D ı2 .Vı Wı /, where each Vı is open in X , each Wı is open in Y , and each
Vı Wı U for some .
S
S
Let x 2 X . Then fxg Y ı2W x2Vı Vı Wı . In particular, Y D ı2W x2Vı Wı . Since Y
S
is compact, 9ı1 ; : : : ; ım 2 , where x 2 Vıi for each i , such that Y D m
i D1 Wıi .
Y
Vı Wı
x
X
T
Sm
Now let Vx ´ m
i D1 Vıi , an open neighbourhood of x satisfying Vx Y i D1 Vıi Wıi . Considering all x 2 X , weSsee the Vx ’s form an open
S cover of X . Since X is compact,
9x1 ; : : : ; xn 2 X such that X D jnD1 Vxj . So X Y D jnD1 Vxj Y .
Each Vxj Y has a finite cover by the Vı Wı ’s, so the same is true for X Y . Using
Vı Wı U , we see that X Y has a finite cover by the U ’s, hence the result.
Remark. Given topological spaces X; Y; Z, the product X Y Z is homeomorphic to
X .Y Z/ (the open sets correspond). By induction, the above theorem implies the
product of finitely many compact spaces is compact.
As a consequence, and with the use of Corollary 4.8 on page 22, Œ M; M n is a compact
subset of Rn . Now a set X Rn is bounded means 9M > 0 such that X Œ M; M n .
So the proof of the same corollary may be extended to show X Rn is compact , X is
closed and bounded.
Proposition 4.12. Let X be a compact metric space, Y be any metric space,
and f W X ! Y be a continuous map. Then f is uniformly continuous, i.e.
8 > 0, 9ı > 0 such that 8x; y 2 X, dX .x; y/ < ı ) dY .f .x/; f .y// < .
Proof. Since f is continuous, 8x 2 X , 9ıx > 0 such that dX .x; x 0 / < 2ıx ) dY .f .x/;f .x 0 //
< 2 . Let Ux ´ fx 0 2 X W dX .x; x 0 / < ıx g. Considering all x 2 X , we
Snsee the Ux ’s form an
open cover of X. Since X is compact, 9x1 ; : : : ; xn 2 X such that X D i D1 Uxi .
Let ı ´ min fıxi g. Suppose dX .y; z/ < ı for some y; z 2 X. Since the Uxi form an open
cover of X , 9 i 2 f1; : : : ; ng such that d.y; xi / < ıxi . Since dX .y; z/ < ı 6 ıxi , we have
dX .z; xi / < 2ıxi by the triangle inequality. Therefore dY .f .y/; f .z// 6 dY .f .y/; f .xi // C
dY .f .xi /; f .z// < 2 C 2 D .
24
4.2
S E Q U E N T I A L C O M PA C T N E S S
4.2 Sequential compactness
Definition 4.13. A topological space X is sequentially compact if every
sequence in X has a convergent subsequence.
For a general topological space, compactness and sequential compactness do not imply each
other. However:
Proposition 4.14. Compact metric spaces are sequentially compact.
Proof. Let .X; d / be a metric space and .xn / be a sequence in X with no convergent subsequence. Then .xn / has infinitely many distinct members. 8x 2 X , 9ı > 0 such that d.x; xn / < ı
for finitely many n only. (If not, then 9x 2 X such that 8m 2 N, d.x; xn / < m1 for infinitely
many n, and so 9 a subsequence of .xn / tending to x.)
We let Ux ´ fy 2 X W d.x; y/ < ıg. Each Ux contains finitely many xn ’s only, so
fUx W x 2 X g is an open cover of X with no finite subcover. X is not compact.
This result implies the Bolzano-Weierstrass theorem: Any closed bounded subset of Rn is
sequentially compact.
Example. Let X Rn be sequentially compact. Then (1) X is bounded: Otherwise 9.xn /
such that d.x; xn / > n, with no convergent subsequence; (2) X is closed: otherwise 9.xn /
such that xn ! x … X .
x 2X nX
X
xn
x1 x2
In fact, we have the following more general result:
Theorem 4.15. Let .X; d / be a sequentially compact metric space. Then:
S
(i) 8 > 0, 9x1 ; : : : ; xn 2 X such that X D niD1 B.xi ; /;
(ii) 8open cover U of X , 9 > 0 such that 8x 2 X, B.x; / U for
some U 2 U;
(iii) .X; d / is compact.
Proof. (i): Suppose not. Then by induction we can construct a sequence .xn / in X such that
d.xm ; xn / > 8m ¤ n. Its subsequences are not Cauchy, hence divergent. Contradiction.
(ii): Suppose not. Then 9 an open cover U of X such that 8n, 9xn 2 X such that B xn ; n1 ›
U , 8U 2 U. But .xn / has a subsequence .xn.r/ / tending
to x 2 X . So let x 2 U0 for some
U0 2 U. Since U0 is open, 9m > 0 such that B x; m2 U0 .
1
Now
9N
such
that
x
n.r/ 2 B x; m 8r > N . Additionally if n.r/ > m, and y 2
1
2
1
B xn.r/ ; n.r/ , then d.x; y/ 6 d.x; xn.r/ / C d.xn.r/ ; y/ < m . So for such n.r/, B xn.r/ ; n.r/
B x; m2 U0 . Contradiction. (See the diagram.)
25
4.2
S E Q U E N T I A L C O M PA C T N E S S
1
B xn.r/ ; n.r/
B x; m1
B x; m2
xn.r/
x
(iii): Let U be an open cover of S
X . Choose an > 0 as provided by (ii). For this , using
(i) 9x1 ; : : : ; xn 2 X suchSthat X D niD1 B.xi ; /. For each i, B.xi ; / Ui for some Ui 2 U
because of (ii). So X D niD1 Ui . X is compact.
26